Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 142 / 3

All Topological K-theory groups below are complex and are indexed modulo two using Bott periodicity. For $n\ge 1$, Complex K-theory of a sphere givesFor an even-dimensional sphere, choose a generator $u$ of Reduced topological K-theory. Its square vanishes: the Chern character sends $u$ to a top-degree cohomology class, whose square is zero, and the Chern character is injective on this torsion-free abelian group. Thus the ring answers areThe exceptional zero-dimensional sphere consists of two points: $K^0(S^0)=\mathbb{Z}\times \mathbb{Z}$ with coordinatewise multiplication and $K^{-1}(S^0)=0$. In particular its reduced generator is idempotent rather than square-zero.

For the Euler characteristic identity, induct over the cells of a finite CW complex. The starting CW complex consisting of $v$ vertices has $K^0\cong {\mathbb{Z}}^v$, $K^{-1}=0$, and $\chi =v$. If $X$ is obtained from $A$ by attaching one $d$-dimensional cell, the quotient is $S^d$, and the K-theory six-term exact sequence isThese abelian groups are finitely generated by induction. Tensoring the exact sequence with $\mathbb{Q}$ preserves exactness. The alternating sum of dimensions in a cyclic six-term exact sequence is zero, soTherefore

There is a genuine convention issue in the nilpotence assertion. With the usual definition of Reduced topological K-theory as the kernel of restriction to one basepoint, the assertion needs $X$ to be connected. For $X=S^0$ based at its first point, $(0,1)$ lies in ${\tilde{K}}^0(X)$ and satisfies $(0,1{)}^m=(0,1)$ for every $m\ge 1$. Thus it is not a nilpotent element. For an arbitrary finite CW complex, the correct assertion uses the rank map in topological K-theory on every connected component:For a connected space, $I(X)={\tilde{K}}^0(X)$.

Here is an induction proving the corrected assertion. On the vertices, $I(X)=0$. For a positive-dimensional cell attachment $A\subset X$, the ideal $J=\ker (K^0(X)\to K^0(A))$ is square-zero. Indeed, lift two elements of $J$ to relative topological K-theory $K^0(X,A)$. Their product is induced by the reduced diagonalThis map is null-homotopic, since $d<2d$ and $S^{2d}$ is $(2d-1)$-connected. The relative product in topological K-theory is therefore zero, giving $J^2=0$. If $a\in I(X)$, its restriction belongs to $I(A)$, so induction gives $(a{|}_A{)}^m=0$ for some $m$. Then $a^m\in J$ and $a^{2m}=0$. We have proved every Topological K-theory class of rank zero on every connected component is a nilpotent element, and hence the requested result for connected $X$.

For the K-theory transfer of a finite covering, let $V\to Y$ be a vector bundle and defineOn an evenly covered neighborhood the $n$ sheets identify this with a direct sum of $n$ local vector bundles. On overlaps the identifications permute the sheets and apply the transition maps of $V$, so these local descriptions glue to a vector bundle on $X$. This operation respects direct sums, hence extends to a homomorphism of Grothendieck groupsIt obeys the projection formula for the K-theory transfer:For actual vector bundles, this follows fiberwise by distributing the tensor product of vector bundles over the direct sum; it then extends to their Grothendieck groups.

Let $P=p_{!}(1_Y)$ be the permutation vector bundle of the covering. It has rank $n$ on every connected component, but need not be a trivial vector bundle; in particular one cannot replace $p_{!}{p}^{\ast }$ by multiplication by $n$ integrally. Instead writeThe componentwise nilpotence result from the preceding solution gives ${\eta }^N=0$ for some $N$. In the localization of a ring $K^0(X)\otimes \mathbb{Z}[1/n]$, the finite geometric seriesis an inverse. The projection formula for the K-theory transfer givesThus $p^{\ast }(a)=0$ implies $a=0$ after inverting $n$. More explicitly, $(n+\eta {)}^{-1}{p}_{!}$ is a left inverse, proving